bzoj 1604: [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居(切比雪夫距离+multiset贪心+并查集)
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1604: [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 1092 Solved: 441
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Description
了解奶牛们的人都知道,奶牛喜欢成群结队.观察约翰的N(1≤N≤100000)只奶牛,你会发现她们已经结成了几个“群”.每只奶牛在吃草的时候有一个独一无二的位置坐标Xi,Yi(l≤Xi,Yi≤[1..10^9];Xi,Yi∈整数.当满足下列两个条件之一,两只奶牛i和j是属于同一个群的:
1.两只奶牛的曼哈顿距离不超过C(1≤C≤10^9),即lXi - xil+IYi - Yil≤C.
2.两只奶牛有共同的邻居.即,存在一只奶牛k,使i与k,j与k均同属一个群.
给出奶牛们的位置,请计算草原上有多少个牛群,以及最大的牛群里有多少奶牛
Input
第1行输入N和C,之后N行每行输入一只奶牛的坐标.
Output
仅一行,先输出牛群数,再输出最大牛群里的牛数,用空格隔开.
Sample Input
4 2
1 1
3 3
2 2
10 10
Sample Output
2 3
和http://blog.csdn.net/jaihk662/article/details/77918172这题处理方法一样,贪心+multiset
①考虑将曼哈顿距离转化为切比雪夫距离
对于点(x, y),令(x', y') = (x+y, x-y),那么有|x1-x2|+|y1-y2|=max(|x1'-x2'|, |y1'-y2'|)<=C
②然后将所有点按x排序,依次加入multiset中,中间保证multiset中任意两点都满足|xi'-xj'|<=C(其中multiset的顺序为y从小到大),这样就只用考虑|y'-yi'|是否<=C了,因为multiset中y有序,所以可以直接二分
④对于第k个点(xk, yk),将它加入multiset之后判断与它相邻的两个点是否满足|yk'-yi'|<=C,满足就并查集并一起
⑤所有点遍历完毕之后就可以统计答案了
#include<stdio.h>#include<set>#include<algorithm>using namespace std;typedef struct Res{int id;int x, y;bool operator < (const Res &b) const{if(y<b.y)return 1;return 0;}}Res;multiset<Res> st;Res s[100005];int ufs[100005], sum[100005];int Find(int x){if(ufs[x]==0)return x;return ufs[x] = Find(ufs[x]);}bool comp(Res a, Res b){if(a.x<b.x)return 1;return 0;}int main(void){int C, n, i, p, t1, t2, ans, cnt;multiset<Res>::iterator it;scanf("%d%d", &n, &C);for(i=1;i<=n;i++){s[i].id = i;scanf("%d%d", &s[i].x, &s[i].y);s[i].x += s[i].y;s[i].y = s[i].x-2*s[i].y;}sort(s+1, s+n+1, comp);p = 1;st.insert(s[1]);for(i=2;i<=n;i++){while(s[p].x+C<s[i].x && p<i){st.erase(st.find(s[p]));p++;}it = st.lower_bound(s[i]);if(it!=st.end() && (*it).y-C<=s[i].y){t1 = Find((*it).id);t2 = Find(s[i].id);if(t1!=t2)ufs[t1] = t2;}if(it!=st.begin() && (*(--it)).y+C>=s[i].y){t1 = Find((*it).id);t2 = Find(s[i].id);if(t1!=t2)ufs[t1] = t2;}st.insert(s[i]);}ans = cnt = 0;for(i=1;i<=n;i++)sum[Find(i)]++;for(i=1;i<=n;i++){if(sum[i])cnt++;ans = max(ans, sum[i]);}printf("%d %d\n", cnt, ans);return 0;}
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